Problem: $ f(x)=\sum_{n=0}^{\infty }\dfrac{n+1}{{{3}^{n+2}}}x^n$ $\int\limits_{0}^{1}f(x)dx=$
Solution: First, perform the requested integration. $\begin{aligned} &\phantom{=}\int\limits_{0}^{1}{f\left( x \right)dx} \\\\ &=\int_0^1\sum_{n=0}^{\infty }\dfrac{n+1}{{{3}^{n+2}}}x^ndx \\\\ &=\sum_{n=0}^{\infty }\int_0^1\dfrac{n+1}{{{3}^{n+2}}}x^ndx \\\\ &=\left. \sum_{n=0}^{\infty }\dfrac{n+1}{{{3}^{n+2}}}\dfrac{x^{n+1}}{n+1} \right|_{0}^{1} \\\\ &=\sum_{n=0}^{\infty }{\dfrac{1}{{{3}^{n+2}}}} \end{aligned}$ Now recognize that the result is a geometric series with first term $\dfrac{1}{9}$ and common ratio $\dfrac{1}{3}$. Hence, the series converges to $\dfrac{\dfrac{1}{9}}{1-\dfrac{1}{3}}=\dfrac{1}{9-3}=\dfrac{1}{6}$